Carbons in a propane sample

Today in class we looked at a problem that some of you didn't quite get to the end of by the time class ended. Here it is. If you haven't already worked it through, give it a good try before you jump ahead to the answer...

How many carbon atoms are in a 37.43L sample of propane gas at 17.52°C and 1.472atm?

This starts out as an Ideal Gas Law problem with a single set of conditions.
PV = nRT
Plugging in the values from the problem:
(1.472atm)(37.43L) = n(0.08206L.atm/mol.K)((17.52+273.15)K)
n = 2.3099mols of C3H8(g)
{NOTE: I'm in the middle of the problem, so I'm not rounding for significant figures yet, 
but it looks like 4 sig figs would be good at this point…}
Each propane molecule contains 3 carbon atoms, so each mole of propane molecules contains 3 moles of carbon atoms.
(2.3099mols C3H8) (3mols C/1mol C3H8) = 6.92974mols C
(6.92974mols C) (6.022x1023 C atoms/mol C) = 4.173x1024 C atoms in the sample.

DISCLAIMER: This problem assumes that propane is behaving as an ideal gas under these conditions. There's a pretty good chance that it would not be ideal under these conditions, but for the purposes of this problem let's assume it is.

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